Abaseball of mass m1 = 0.28 kg is thrown at another ball hanging from the ceiling by a length of string l = 1.35 m. the second ball m2 = 0.92 kg is initially at rest while the baseball has an initial horizontal velocity of v1 = 3.5 m/s. after the collision the first baseball falls straight down (no horizontal velocity.)

The angle is 16.86 °

Explanation:

Step 1: Given Data

mass of the baseball 1 = 0.28 kg

lengt of string L = 1.35m

mass of ball 2 = 0.92 kg

horizontal velocity = 3.5 m/s

After the collision the first baseball falls straight down (no horizontal velocity).

Step 2: Calculate the velocity

The linear moment before the shock is equal to the linear moment after the shock:

m 1 *u 1 + 0 = 0 + m2*v

v= u1 * m1/m2  = 3.5m/S * ( 0.28/0.92)

v = 1.065 m/s

Step 3: Calculate the height

The kinetic energy of ball 2, immediately after the shock, is equal to the final gravitational potential energy

Kinitial = Ug, final

1/2*m2*v² = ms * g*h

h= v²/2g

h= 1.065²/(2*9.81)

h = 0.058 meter

Step 4: Calculate the maximum angle the rope makes with the vertical is:

cos(∅) = (L-h)/L

∅ = cos^-1 ((1.35-0.058)/1.35)

∅ = cos^-1(0.957)

∅ =16.86°

The angle is 16.86 °

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr